Q: What decimal number does the binary number 1000 0000 represent?

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The decimal equivalent of the binary number 10110000 is 176

128

Binary for the decimal number 2 is 10 (or 0000 0010 if you want it as a byte value) ASCII code for the decimal number 2 is 50 (Dec), 34 (Hex), 062 (Oct), 0011 0010 (Bin)

At least 8 bits are needed to represent the number 231. 231 ÷ 2 = 115 r 1 115 ÷ 2 = 57 r 1 57 ÷ 2 = 28 r 1 28 ÷ 2 = 14 r 0 14 ÷ 2 = 7 r 0 7 ÷ 2 = 3 r 1 3 ÷ 2 = 1 r 1 1 ÷ 2 = 0 r 1 → 231 is 1110 0111 in binary and has 8 binary digits. Thus 231 can be represented in 8 bits, but if more are provided, eg 16, it can still be represented (in 16 bits it would be 0000 0000 1110 0111, unless there is a binary point, with say 8 bits after it, then 231 would be 1110 0111 . 0000 0000).

Convert each binary number to decimal: Each digit to the left is worth twice what it is worth in the column to its right: each column starting from the right is worth 1, 2, 4, 8, 16, 32, ...; for an 8-bit number, the columns from the left are 128, 64, 32, 16, 8, 4, 2, 1; thus : 0000 1010 = 8 + 2 = 10 0110 0100 = 64 + 32 + 4 = 100 0001 0101 = 16 + 4 + 1 = 21 0000 0001 = 1 → 00001010.01100100.00010101.00000001 = 10.100.21.1 The IPv4 dotted decimal representation is just expressing the 32 bits of the address in the value of each of the 4 bytes used to store the 32 bits using decimal numbers with a dot between each byte's value so that it is not confused to be a single number, ie so that 10.100.21.1 is not read as 10100211 or 10,100,021,001 when the correct decimal value of the 32 bit binary number (0000 1010 0110 0100 0001 0101 0000 0001) is 174,331,137 (but converting that to binary is not as easy as converting an 8 bit (decimal number of range 0-255) to binary.

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The decimal equivalent of the binary number 10110000 is 176

To consider the difference between straight binary and BCD, the binary numbers need to be split up into 4 binary digits (bits) starting from the units. In 4 bits there are 16 possible values from 0000 to 1111 (0 to 15). In straight binary all of these possible combinations are used, thus: 4 bits can represent the decimal numbers 0-15 8 bits can represent the decimal numbers 0-255 12 bits can represent the decimal numbers 0-4095 16 bits can represent the decimal numbers 0-65535 etc In arithmetic, all combinations of bits are used, thus: 0000 1001 + 0001 = 0000 1010 In BCD or Binary Coded Decimal, only the representations of the decimal numbers 0-9 are used (that is 0000 to 1001 in binary), and the 4-bits (nybbles) are read as decimal digits, thus: 4 bits can represent the decimal digits 0-9 8 bits can represent the decimal digits 0-99 12 bits can represent the decimal digits 0-999 16 bits can represent the decimal digits 0-9999 In arithmetic, only the representations of decimal numbers are used, thus: 0000 1001 + 0001 = 0001 0000 When BCD is used each half of a byte is read directly as a decimal digit. BCD is obviously inefficient as storage (for large numbers) as each nybble is only holding 3/8 of the possible numbers, however, it is sometimes easier and quicker to work with decimal digits (for example when there is lots of display of counting numbers to do there is less binary to decimal conversion needing to be done).

I assume you mean BCD, Binary Coded Decimal. BCD uses 4 bits to represent one decimal number. The easiest way is to make a table, with decimal, BCD, Hex and straight binary. 1 0000 0001 1 0000 0001 2 0000 0010 2 0000 0010 3 0000 0011 3 0000 0011 ...Skip a bit.... 9 0000 1001 9 0000 1001 10 0001 0000 A 0000 1010 11 0001 0001 B 0000 1011 ...Skipping again.... 15 0001 0101 F 0000 1111 16 0001 0110 10 0001 0000 Get the idea? In the first one, 4 binary bits are matched with one decimal digit. In straight binary, the number scrolls on. Interestingly, this caused some problems, earning itself the name 'the 2.1K bug'. some systems, generally small systems like Eftpos terminals, wrote values in BCD binary, but read them as straight binary. So dates were written in BCD 10, but read back as (check the table) Ordinary binary 16. Hilarity ensued.

When a bit is turned on, it represents a "1". When it is turned off, it represents a "0". The exact value depends on where the bit is within the byte it is part of. In the binary number 0000 0001, the last bit is set to 1 and represents the number 1. In the binary number 0000 0010, the second to last bit is set to 1, which corresponds to the "2's" place relative to decimal numbers. In the binary number 0000 1000, the bit that is set to 1 represents the value "8" in decimal numbers.

Decimal 30 = binary 11110. The decimal binary code (BCD), however, is 11 0000.

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00001001 can be written as 0000 1001 which is hex 09 and hence has a decimal value of 9

In decimal, 4 digits ( 0 - 9) you can represent any number from 0000 to 9999. That is 10,000 unique number combinations. Incidentally, the number of unique cominations is 10 to the 4th power. In binary, 4 digits (0 - 1) can represent 16 different cominations from 0000 - 1111. This is given by 2 to the 4th power

Binary for the decimal number 2 is 10 (or 0000 0010 if you want it as a byte value) ASCII code for the decimal number 2 is 50 (Dec), 34 (Hex), 062 (Oct), 0011 0010 (Bin)

Floating point numbers are stored in scientific notation using base 2 not base 10.There are a limited number of bits so they are stored to a certain number of significant binary figures.There are various number of bytes (bits) used to store the numbers - the bits being split between the mantissa (the number) and the exponent (the power of 10 (being in the base of the storage - in binary, 10 equals 2 in decimal) by which the mantissa is multiplied to get the binary/decimal point back to where it should be), examples:Single precision (IEEE) uses 4 bytes: 8 bits for the exponent (encoding ±), 1 bit for the sign of the number and 23 bits for the number itself;Double precision (IEEE) uses 8 bytes: 11 bits for the exponent, 1 bit for the sign, 52 bits for the number;The Commodore PET used 5 bytes: 8 bits for the exponent, 1 bit for the sign and 31 bits for the number;The Sinclair QL used 6 bytes: 12 bits for the exponent (stored in 2 bytes, 16 bits, 4 bits of which were unused), 1 bit for the sign and 31 bits for the number.The numbers are stored normalised:In decimal numbers the digit before the decimal point is non-zero, ie one of {1, 2, ..., 9}.In binary numbers, the only non-zero digit is 1, so *every* floating point number in binary (except 0) has a 1 before the binary point; thus the initial 1 (before the binary point) is not stored (it is implicit).The exponent is stored by adding an offset of 2^(bits of exponent - 1), eg with 8 bit exponents it is stored by adding 2^7 = 1000 0000Zero is stored by having an exponent of zero (and mantissa of zero).Example 10 (decimal):10 (decimal) = 1010 in binary → 1.010 × 10^11 (all digits binary) which is stored in single precision as:sign = 0exponent = 1000 0000 + 0000 0011 = 1000 00011mantissa = 010 0000 0000 0000 0000 0000 (the 1 before the binary point is explicit).Example -0.75 (decimal):-0.75 decimal = -0.11 in binary (0.75 = ½ + ¼) → 1.1 × 10^-1 (all digits binary) → single precision:sign = 1exponent = 1000 0000 + (-0000 0001) = 0111 1111mantissa = 100 0000 0000 0000 0000 0000Note 0.1 in decimal is a recurring binary fraction 0.1 (decimal) = 0.0001100110011... in binary which is one reason floating point numbers have rounding issues when dealing with decimal fractions.

The 1's complement is formed by inverting every binary digit (bit) of the number - if it is a 0 it becomes a 1, otherwise it is a 1 and becomes a 0. If 10 is in base 2, then its 1's compliment is 01 or just 1. If 10 is in base 10, then in binary it is 1010 and its 1's complement is 0101 = 5 in decimal. However, if more bits are being used to store it, there would be leading 0s that get inverted to 1s and so the resultant number is different; examples: 8 bits (a byte): decimal 10 = 0000 1010 → 1111 0101 = 245 in decimal 16 bits: decimal 10 = 0000 0000 0000 1010 → 1111 1111 1111 0101 = 65525 Next, if 2s complement is being used to represent negative numbers, the binary 1111 0101 represents decimal -11; similarly 1111 1111 1111 0101 represents decimal -11.

At least 8 bits are needed to represent the number 231. 231 ÷ 2 = 115 r 1 115 ÷ 2 = 57 r 1 57 ÷ 2 = 28 r 1 28 ÷ 2 = 14 r 0 14 ÷ 2 = 7 r 0 7 ÷ 2 = 3 r 1 3 ÷ 2 = 1 r 1 1 ÷ 2 = 0 r 1 → 231 is 1110 0111 in binary and has 8 binary digits. Thus 231 can be represented in 8 bits, but if more are provided, eg 16, it can still be represented (in 16 bits it would be 0000 0000 1110 0111, unless there is a binary point, with say 8 bits after it, then 231 would be 1110 0111 . 0000 0000).